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Câu hỏi: Cho hàm số $f(x)$ có đạo hàm liên tục trên $\left[ 0;1 \right]$ thỏa mãn $f(1)=4;f(0)=1$ và $\int_{0}^{1}{{{\left[ {{f}^{\prime }}(x) \right]}^{2}}}dx=9$. Giá trị cùa tích phân $\int_{0}^{1}{x}\cdot {{f}^{2}}(x)dx$ bằng
A. $\dfrac{1}{4}$.
B. $9$.
C. $\dfrac{1}{6}$.
D. $\dfrac{19}{4}$.
A. $\dfrac{1}{4}$.
B. $9$.
C. $\dfrac{1}{6}$.
D. $\dfrac{19}{4}$.
Ta có $I=\int_{0}^{1}{{{f}'}}(x)dx=f\left( 1 \right)-f\left( 0 \right)=4-1=3.$
$\int_{0}^{1}{3}dx=3\Rightarrow {{\left( \int_{0}^{1}{3}dx \right)}^{2}}=9.$
$\int_{0}^{1}{{{\left( {{f}^{\prime }}(x) \right)}^{2}}}dx-2.3.\int_{0}^{1}{{f}'\left( x \right)}dx+{{\left( \int_{0}^{1}{3}dx \right)}^{2}}=9-2.3.3+{{3}^{2}}=0\Rightarrow \int_{0}^{1}{{{\left( {{f}^{\prime }}(x)-3 \right)}^{2}}}dx=0$ $\Rightarrow {f}'(x)-3=0$
$\Rightarrow {{f}^{\prime }}(x)=3\Rightarrow f\left( x \right)=3x+C.$
Theo bài ra $f(0)=1\Rightarrow C=1\Rightarrow f\left( x \right)=3x+1\Rightarrow \int\limits_{0}^{1}{x.{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{1}{x.{{\left( 3x+1 \right)}^{2}}dx}=\dfrac{19}{4}.$
$\int_{0}^{1}{3}dx=3\Rightarrow {{\left( \int_{0}^{1}{3}dx \right)}^{2}}=9.$
$\int_{0}^{1}{{{\left( {{f}^{\prime }}(x) \right)}^{2}}}dx-2.3.\int_{0}^{1}{{f}'\left( x \right)}dx+{{\left( \int_{0}^{1}{3}dx \right)}^{2}}=9-2.3.3+{{3}^{2}}=0\Rightarrow \int_{0}^{1}{{{\left( {{f}^{\prime }}(x)-3 \right)}^{2}}}dx=0$ $\Rightarrow {f}'(x)-3=0$
$\Rightarrow {{f}^{\prime }}(x)=3\Rightarrow f\left( x \right)=3x+C.$
Theo bài ra $f(0)=1\Rightarrow C=1\Rightarrow f\left( x \right)=3x+1\Rightarrow \int\limits_{0}^{1}{x.{{f}^{2}}\left( x \right)dx}=\int\limits_{0}^{1}{x.{{\left( 3x+1 \right)}^{2}}dx}=\dfrac{19}{4}.$
Đáp án D.